Q. An electric dipole is placed at an angle of $60^{0}$ with an electric field of intensity 10$^{5}$ NC$^{-1}$. It experiences a torque equal to $8\sqrt{3}$ Nm. Calculate the charge on the dipole, if the dipole length is 2 cm.
A) $- 8 \times 10^{3\ }C$
B) $8.54 \times 10^{- 4\ }C$
C) $8 \times 10^{- 3\ }C$
D) $0.85 \times 10^{- 6\ }C$
Explanation
(c) Here, $\theta = 60^{0},E = 10^{5}{NC}^{- 1}$
$\tau = 8\sqrt{3\ }Nm,q = ?,2a = 2cm = 2 \times 10^{- 2}m$
From $\ \ \tau = pE\sin{\theta = q(2a)}E\sin\theta$
$q = \frac{\tau}{2aE\sin\theta}\ = {\ \frac{8\sqrt{3}}{2 \times 10^{- 2} \times 10^{5} \times {sin60}{^\circ}^{\ }}}$
$= \frac{8\sqrt{3}}{2 \times 10^{3} \times \sqrt{3}/2}\ $
$q = 8 \times 10^{- 3\ }C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.